Question- Triangle $ABC$ has incenter $I$. Let points $X$, $Y$ be located on the line segments $AB$, $AC$ respectively, so that $BX\cdot AB = IB^2$ and $CY\cdot AC = IC^2$. Given that the points $X, I, Y$ lie on a straight line, find the possible values of the measure of angle $A$?
I have solved it and my answer is $60^{\circ}$. Are there any other possible values of $A$, because the question says "possible values of $A$"? Have I missed anything or any other values of $A$?
Please help. And this question is from the Indian National Mathematics Olympiad - 1991.
By the given $BI$ is a tangent line to the circle $(AIX)$, which gives $\measuredangle XIB=\frac{\alpha}{2}$.
Similarly, $\measuredangle YIC=\frac{\alpha}{2}.$
Thus, since $X$, $Y$ and $I$ are collinear, we obtain: $$\alpha=\frac{\beta}{2}+\frac{\gamma}{2}$$ or $$2\alpha=180^{\circ}-\alpha$$ or $$\alpha=60^{\circ}.$$